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(A) Let $OA = x \ km$,$OB = y \ km$,and $AB = R \ km$.Using the Law of Cosines in $	riangle AOB$:$R^2 = x^2 + y^2 - 2xy \cos(120^o)$Since $\cos(120^o

Vedclass · Accepted Answer

$\frac{260}{\sqrt{37}}$

Vedclass · Answer

(A) Let $OA = x \ km$,$OB = y \ km$,and $AB = R \ km$.Using the Law of Cosines in $	riangle AOB$:$R^2 = x^2 + y^2 - 2xy \cos(120^o)$Since $\cos(120^o) = -\frac{1}{2}$,we have:$R^2 = x^2 + y^2 - 2xy(-\frac{1}{2}) = x^2 + y^2 + xy \quad \dots(1)$At the given instance,$x = 8 \ km$ and $y = 6 \ km$:$R^2 = 8^2 + 6^2 + (8 	imes 6) = 64 + 36 + 48 = 148$$R = \sqrt{148} = 2\sqrt{37} \ km$.Differentiating equation $(1)$ with respect to time $t$:$2R \frac{dR}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} + (x \frac{dy}{dt} + y \frac{dx}{dt})$Given $\frac{dx}{dt} = 20 \ km/hr$ and $\frac{dy}{dt} = 30 \ km/hr$:$2(2\sqrt{37}) \frac{dR}{dt} = 2(8)(20) + 2(6)(30) + (8 	imes 30 + 6 	imes 20)$$4\sqrt{37} \frac{dR}{dt} = 320 + 360 + (240 + 120)$$4\sqrt{37} \frac{dR}{dt} = 680 + 360 = 1040$$\frac{dR}{dt} = \frac{1040}{4\sqrt{37}} = \frac{260}{\sqrt{37}} \ km/hr$.

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